# Operational Amplifier with mixed feedback

So, I came across this circuit in a test, it's regarding the operation of an op amp with multiple (positive and negative) feedback, the circuit is as follows: 1. Does the Operational Amplifier operates as a hysteresis comparator due to the positive feedback formed by the R3, R4, R5 and RL?

Since it's an ideal op amp and the Vp and Vn supply voltage wasn't informed I assumed this is not the case, and also in regular hysteresis comparators you usually only have a positive feedback instead of mixed.

2. The transconductance gain of the circuit is $$\\frac{I_{o}}{V_{i}}= -1\frac{mA}{V}\$$

Here's where I'm having some trouble. I've simulated the circuit and indeed this seems to be the case, but I'm having difficulty to prove it with math.

I know that in an ideal op amp, the current entering the inverting and non-inverting nodes are zero and the node voltages at the input nodes are equal.

With that in mind, and starting with the inverting ouput, I suppose we can calculate the input current as follows:

$$\I_{i}=\frac{\left ( V_{i}-V_{a} \right )}{R1+R2}\$$

Where $$\V_{a}\$$ is the voltage at the output node of the op amp.

Since there's no current flowing through the inverting and non-inverting nodes, we can consider that $$\R3\$$ and $$\R4\$$ are in series, right? But what about the voltage $$\V+\$$? I guess I should apply KCL on the non-inverting node as well but I'm having difficulty understanding how exacly am I supposed to do that.

3. The circuit is linear and the ouput voltage $$\V_{o}\$$ is given by $$\V_{o}=RL\cdot I_{o}\$$

This seems like a no-brainer to me, because assuming that the op-amp in operating in the linear region the ouput is simply going to be defined by the current across $$\RL\$$.

So I was hoping someone can help me with the circuit analysis, specially item 2.

Edit:

So, before I saw the TimWescott answer I tried to solve it without superposition or other simplification method and here's what I got:

I defined the node at the op amp as Va and tried to apply KCL:

• Inverter node, solving for $$\V_{a}\$$ and then replacing the $$\R\$$ values and simplifying:

$$\\frac{V_{a}-V_{i}}{R_{1}}+\frac{V_{a}-V_{out}}{R_{2}}=0 \therefore V_{a}=\frac{R_{1}\cdot V_{out}+R_{2}\cdot V_{i}}{R_{1}+R_{2}} \therefore V_{a}=\frac{2\cdot V_{out}+V_{i}}{3}\$$

• Non-inverting node, solving for $$\V_{a}\$$ and then replacing the $$\R\$$ values and simplifying:

$$\\frac{V_{a}-0}{R_{3}}+\frac{V_{a}-V_{o}}{R_{4}}=0 \therefore V_{a}=\frac{R_{3}\cdot V_{o}}{R_{3}+R_{4}} \therefore V_{a}\cong \frac{2}{3}\cdot V_{o}\$$

• Equating and solving for $$\V_{out}\$$:

$$\\frac{2\cdot V_{out}+V_{i}}{3}=\frac{2}{3}\cdot V_{o}\therefore V_{out}=V_{o}-\frac{1}{2}\cdot V_{i}\$$ (1)

• $$\V_{o}\$$ node:

$$\\frac{V_{o}-V_{out}}{R_{5}}+\frac{V_{o}-V_{a}}{R_{4}}+I_{o}=0\$$ (2)

• $$\V_{o}\$$ as function of $$\I_{o}\$$:

$$\V_{o}=I_{o}\cdot RL \therefore V_{o}=I_{o}\cdot 100\$$ (3)

• Finally, replacing (3) in (1) and then both in (2) and solving for Io (a lot of math involved lol):

$$\I_{o}\cong -V_{i}\cdot 9.993\cdot 10^{-4} \therefore I_{o}\approx -V_{i}\cdot 1\cdot 10^{-3}\$$

And therefore:

$$\\frac{I_{o}}{V_{i}}\approx -1 \frac{mA}{V}\$$

Well that makes statement 2 right indeed.

But it also makes statement 3 correct doesn't it? I actually used that very equation to find the "transconductance gain".

Also, I can determine the voltage gain from the previous math:

$$\\frac{V_{o}}{R_{L}}\approx -V_{i}\cdot 1\cdot 10^{-3}\therefore \frac{V_{o}}{100}\approx -V_{i}\cdot 1\cdot 10^{-3}\therefore \frac{V_{o}}{V_{i}}\approx -0.1\$$

Which appears to check out as well (as Sunnyskyguy EE75 answered for instance).

I also tried to follow TimWescott tips, but I'm not sure I got it right:

• Setting $$\V_{o}\$$ to zero and solving for $$\V_{out}\$$ as a function of $$\V_{i}\$$:

$$\\frac{V_{a}-V_{i}}{R_{1}}+\frac{V_{a}-V_{out}}{R_{2}}=0\$$ and $$\\frac{V_{a}-0}{R_{3}||R_{4}}=0 \therefore V_{a}=0\$$

Hence:

$$\V_{out}=-\frac{R_{2}}{R_{1}}\cdot V_{i}\$$

• Setting $$\V_{i}\$$ to zero and solving for $$\V_{out}\$$ as a function of $$\V_{o}\$$:

$$\\frac{V_{a}-0}{R_{1}}+\frac{V_{a}-V_{out}}{R_{2}}=0\$$ and $$\\frac{V_{a}-0}{R_{3}}+\frac{V_{a}-V_{o}}{R_{4}}=0\$$

Equating and solving for $$\V_{out}\$$:

$$\V_{out}=\frac{R_{3}\cdot \left ( R_{1}+R_{2} \right )}{R_{1}\cdot \left ( R_{3}+R_{4} \right )}\cdot V_{o}\$$

But if I join the two equations and calculate for gain I get ~0.5 instead of 0.1, I guess I did something wrong.

Also I'm not sure what you meant by "Then use that to calculate $$\I_{o}\$$ as a function of those two things.".

• A comparator circuit would generally have no negative feedback. My suggestion would be to perform a nodal analysis. With an ideal op-amp, you should be able to add your positive feedback signal similar to the output. Sep 17 '19 at 21:06
• It looks kind of like a Howland Current Source. Sep 17 '19 at 21:11
• @Caleb Reister It does, I'm gonna look into that. Sep 17 '19 at 21:24
• You should calculate de Beta (thats the feedback factor) due to negative feedback, and the Beta due to positive feedback, the difference between the 2 gives the actual Beta, if the Beta due to positive feedback is larger, positive feedback wins, if its lower, negative feedback wins, if im not mistaken, in your drawing Beta neg = 0.667, whilst Beta pos = 0.111, so negative feedback wins, and your circuit does not behave as a Schmitt trigger.
– S.s.
Sep 17 '19 at 21:27

First of all, you wrote: "defined by the current across" but the current is trough a component, voltage is across.

Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write for every node the node equation:

$$\begin{cases} \text{I}_{\text{R}_1}+\text{I}_{\text{R}_2}=0\\ \\ \text{I}_\text{op}=\text{I}_{\text{R}_2}+\text{I}_{\text{R}_5}\\ \\ \text{I}_{\text{R}_4}=\text{I}_{\text{R}_3}\\ \\ \text{I}_{\text{R}_5}=\text{I}_{\text{R}_4}+\text{I}_{\text{R}_\text{L}} \end{cases}\tag1$$

Using KVL, we can write for every resistor the voltage-current relation using Ohm's law:

$$\begin{cases} \text{I}_{\text{R}_1}=\frac{\text{V}_\text{i}-\text{V}_-}{\text{R}_1}\\ \\ \text{I}_{\text{R}_2}=\frac{\text{V}_1-\text{V}_-}{\text{R}_2}\\ \\ \text{I}_{\text{R}_3}=\frac{\text{V}_+}{\text{R}_3}\\ \\ \text{I}_{\text{R}_4}=\frac{\text{V}_\text{o}-\text{V}_+}{\text{R}_4}\\ \\ \text{I}_{\text{R}_5}=\frac{\text{V}_1-\text{V}_\text{o}}{\text{R}_5}\\ \\ \text{I}_{\text{R}_\text{L}}=\frac{\text{V}_\text{o}}{\text{R}_\text{L}} \end{cases}\tag2$$

Now, we can put equations $$\(1)\$$ into $$\(2)\$$ to get:

$$\begin{cases} \frac{\text{V}_\text{i}-\text{V}_-}{\text{R}_1}+\frac{\text{V}_1-\text{V}_-}{\text{R}_2}=0\\ \\ \text{I}_\text{op}=\frac{\text{V}_1-\text{V}_-}{\text{R}_2}+\frac{\text{V}_1-\text{V}_\text{o}}{\text{R}_5}\\ \\ \frac{\text{V}_\text{o}-\text{V}_+}{\text{R}_4}=\frac{\text{V}_+}{\text{R}_3}\\ \\ \frac{\text{V}_1-\text{V}_\text{o}}{\text{R}_5}=\frac{\text{V}_1-\text{V}_+}{\text{R}_4}+\frac{\text{V}_\text{o}}{\text{R}_\text{L}} \end{cases}\tag3$$

Now, we can solve for the transfer function because we know that in an ideal opamp we get $$\\text{V}_-=\text{V}_+\$$:

$$\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=\frac{\text{R}_2\text{R}_\text{L}(\text{R}_3+\text{R}_4)}{\text{R}_2\text{R}_3\text{R}_\text{L}-\text{R}_1(\text{R}_5(\text{R}_3+\text{R}_4)+\text{R}_\text{L}(\text{R}_4+\text{R}_5))}\tag4$$

$$\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=-\frac{1}{10}=-0.1\tag5$$

I checked my formula using LTspice and got the same result.

Now, we also get that:

$$\frac{\text{I}_\text{o}}{\text{V}_\text{i}}=\frac{\text{V}_\text{o}}{\text{V}_\text{i}}\cdot\frac{1}{\text{R}_\text{L}}=-\frac{1}{10}\cdot\frac{1}{100}=-\frac{1}{1000}=-0.001\tag6$$

And the input current is defined as:

$$\text{I}_\text{i}=\text{I}_{\text{R}_1}=\frac{\text{V}_\text{i}-\text{V}_-}{\text{R}_1}=\frac{319}{29900000}\cdot\text{V}_\text{i}\approx0.0000106689\cdot\text{V}_\text{i}\tag7$$

In order to make the output current independent of the load resistor, we need to satisfy the following condition:

$$\text{R}_2\text{R}_3-\text{R}_1(\text{R}_4+\text{R}_5)=0\tag8$$

And when we have that condition we get:

$$\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=-\frac{\text{R}_2\text{R}_\text{L}}{\text{R}_1\text{R}_5}\tag9$$

And the output current is then:

$$\text{I}_\text{o}=\frac{\text{V}_\text{o}}{\text{R}_\text{L}}=-\frac{\text{R}_2\text{V}_\text{i}}{\text{R}_1\text{R}_5}\tag{10}$$

• Also, try to show that if you pick $\frac{R_2}{R_1} = \frac{R_4+R_5}{R_3}$ then you get the load current independent of the load resistance. And receive a true VCCS (Voltage controlled current source).
– G36
Sep 18 '19 at 19:48
• @G36 I think that that equation is wrong, I edited my answer and got a different formula.
– Jan
Sep 18 '19 at 20:13
• Try it in LTspice set R1= R3 = R4 = R5 = R and R2 = 2R and you will see that we will have a VCCS.
– G36
Sep 18 '19 at 20:21
• Also, notice that the OP circuit also meats this condition R2/R1 = (R4 +R5)/R3 coincidence?
– G36
Sep 18 '19 at 20:23
• Good job. Thanks Sep 19 '19 at 2:45
1. Right answer, wrong reasons. Just because someone is too lazy to show the power supply rails doesn't mean they're there. You could easily change the values of some of the resistances and make this thing into a comparator with hysteresis.
2. Break the problem down! You have options, but I would set $$\V_o\$$ to zero, and solve for $$\V_{out}\$$ as a function of $$\V_i\$$. Then set $$\V_i\$$ to zero and solve for $$\V_{out}\$$ as a function of $$\V_o\$$. Then use that to calculate $$\I_o\$$ as a function of those two things. You should find that $$\I_o\$$ is independent of $$\V_o\$$.